# Prove that for every $n \in \mathbb{N}$ $\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$ [duplicate]

$$\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$$

1. First step would be proving that the statement is true for n=2

On the LHS for $$n=2$$ we would have $$\frac{1}{4}$$ therefore the statement is true for $$n=2$$

1. Now we must assume the statement is true for $$n=j$$ with $$j\geq2$$

$$\sum\limits_{k=2}^{j}{\frac{1}{k^2}}<1$$

1. Now we must prove true for $$n=j+1$$

$$\sum\limits_{k=2}^{j+1}{\frac{1}{k^2}}<1$$

$$\Rightarrow \sum\limits_{k=2}^{j}{\frac{1}{k^2}}+\frac{1}{(k+1)^2}<1$$

I cant seem to introduce the induction hypothesis

• "Blindly" using induction would clearly not work since the RHS doesn't increase. So what you want to do is "stronger" induction where you strengthen the hypothesis. Apr 26, 2020 at 16:46

Let $$n\in\mathbb{N} :$$

Notice that for any integer $$k\geq 2$$, we have $$k^{2}\geq k\left(k-1\right)$$, and thus : $$\sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\sum_{k=2}^{n}{\frac{1}{k\left(k-1\right)}}=\sum_{k=2}^{n}{\left(\frac{1}{k-1}-\frac{1}{k}\right)}=1-\frac{1}{n}<1$$

• (+1) for the "classical" approach Apr 26, 2020 at 17:03

Strengthen the induction hypothesis

Hint: Prove by induction that

$$\sum_{k=2}^n \frac{1}{k^2 } < 1 - \frac{1}{n}.$$

• Im not sure I quite follow. How is this equivalent to the initial statement? Apr 26, 2020 at 16:54
• They are not equivalent. However, if you prove the inequality of @Calvin Lin, then, you are done. As an analogy, suppose you need to show that $x \leq 3$. Certainly, one way to show this is that $x \leq 2$:) It is the same logic here. Also just as a side note if you use the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$. Then, the result follows, as well. Of course, it is an overkill for this question. Since we are using a fact which is more difficult than the question itself.
– ALNS
Apr 26, 2020 at 16:59
• @MarkViola Thanks, fixed the typo. IMO For "obvious" typos like this, you can go ahead and edit the writeup. Apr 26, 2020 at 17:04

Here is another approach :

We can easily prove that $$\left(\forall x\in\left[0,\frac{1}{2}\right]\right),\ \frac{2}{\pi}\arctan{\left(2x\right)}-x\geq 0$$ (By differentiation the function $$x\mapsto\frac{2}{\pi}\arctan{\left(2x\right)}-x$$, studying its variations, etc...)

Thus, if $$n$$ is a positive integer, we have the following : \begin{aligned}\sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{2}{k^{2}}\right)}}&=\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{\left(k+1\right)-\left(k-1\right)}{1+\left(k+1\right)\left(k-1\right)}\right)}}\\&=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{k}\right)}+\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{k}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\left(\arctan{\left(n+1\right)}-\arctan{2}\right)+\frac{2}{\pi}\left(\arctan{n}-\frac{\pi}{4}\right)\\ &\leq\frac{3}{2}-\frac{2}{\pi}\arctan{2}<1\end{aligned}

To get to the last line we upper bounded the $$\arctan$$s by $$\frac{\pi}{2} \cdot$$

$$f(x)= \frac{1}{x^2}$$, $$x >0$$, stricly decreasing.

$$\displaystyle{\sum_{k=2}^{n}}\frac{1}{k^2} <\displaystyle{ \int_{1}^{n}}\frac{1}{x^2}dx =-\frac{1}{x}\big ]_1^n=$$

$$1-1/n.$$

As suggested by Calvin you can use the above inequality for the induction proof.

• This was already posted 5 minutes ago Apr 26, 2020 at 17:04